Show f(x) = { x/2 if x rational , x if x irrational is not differentiable at 0

[kbgregory]

Homework Statement

Show that the function

f(x)

= { x/2 if x is rational
{ x if x irrational

is not differentiable at 0

Homework Equations

If f is differentiable at 0 then for every e > 0 there exists some d > 0 such that when |x| < d, |(f(x)-f(0))/x - L | < e for some L, which is the derivative of f at 0.

The Attempt at a Solution

Thus far I have:

Choose e = 1/4. Suppose f is differentiable at 0 and let L be f'(0). Then there is some d such that whenever |x| < d,

|(f(x)-f(0))/x - L |

= {|(x/2-0)/x - L | = | .5 - L | if x is rational

{|(x-0)/x - L | = | 1 - L | if x is irrational

Thus we have these inequalities for L:

| .5- L | < .25 and | 1 - L | < .25

which together imply that -.25 < L < .75 as well as that .75 < L < 1.25 which is a contradiction. Therefore f is not differentiable at 0.

Is this correct?

 

[VeeEight]

I have not looked through your work but the usual way to do this problem is to set up the slope at the point (in this case, 0), and show the limit does not exist. That is, work with f(x)/x as x tends to 0.

 

[Office_Shredder]

It looks good. You basically know that there are two slopes: one of .5, and one of 1, and use that information to prove that there cannot be a single slope

 

[HallsofIvy]

You should be looking at (f(x)- f(0))/(x- 0)= f(x)/x. Now look at what happens if x is rational or irrational.

 

[GVAR717]

I think HallsofIvy's process is easier but it seems that either method works. I think the necessary element is that 1/2 does not equal 1.